R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6C

R S Aggarwal Solutions for Class 10 Maths Chapter 6 exercise 6c is available here. These R S Aggarwal solutions are prepared by subject experts to help students clear their doubts. The concepts are explained using step by step approach. This exercise is based on topic area of a triangle using vertex coordinates. Students can download Class 10 Maths Chapter 6 R S Aggarwal Solutions and start practicing offline.

Download PDF of R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6C

RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C
RS Aggarwal Sol class 10 Maths Chapter 6C

 

Access other exercise solutions of Class 10 Maths Chapter 6 Coordinate Geometry

Exercise 6A Solutions: 20 Questions (20 Short Answers)

Exercise 6B Solutions: 20 Questions (20 Short Answers)

Exercise 6D Solutions: 17 Questions (12 Short Answers & 5 Long Answers )

Exercise 6C Page No: 336

Question 1: Find the area of ∆ABC whose vertices are:

(i) A (1, 2), B (-2, 3) and C (-3, -4)

(ii) A (-5, 7), B (-4, -5) and C (4, 5)

(iii) A (3, 8), B (-4, 2) and C (5, -1)

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Solution:

Area of ∆ABC whose vertices are is (x_1,y_1 ),(x_2,y_2 )and (x_3,y_3 ) are

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -4)

Area of triangle = 1/2(1(–2 + 3)–2(–4–2)–3(2–3))

= 1/2(1 + 12 + 3)

= 8 sq units

(ii) A (-5, 7), B (-4, -5) and C (4, 5)

Area of triangle = 1/2(–5(–5–5)–4(5–7) + 4(7 + 5))

= 1/2(–50 + 8 + 48)

= 5 sq units

(iii) A (3, 8), B (-4, 2) and C (5, -1)

Area of triangle = 1/2(3(2 + 1)–4(–1–8) + 5(8–2))

= 1/2(9 + 36 + 30)

= 1/2(75)

= 37.5 sq units

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Area of triangle = 1/2(10(5–3) + 2(3 + 6)–1(–6–5))

= 1/2(20 + 18 + 11)

= 1/2(49)

= 24.5 sq units

Question 2: Find the area of quadrilateral ABCD whose vertices are A (3, -1), B (9, -5), C (14, 0) and D (9, 19).

Solution: Vertices of quadrilateral ABCD are A(3,-1), B(9, -5), C (14, 0) and D(9, 19)

Construction: Join diagonal AC.

We know that:

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 1

Question 3: Find the area of quadrilateral PQRS whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2).

Solution:

Given: PQRS is a quadrilateral whose vertices are P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2)

Construction: Join PR

We know that:

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 2

Now, Area of quadrilateral PQRS = Area of triangle PQR + Area of triangle PSR

= 21/2 + 35/2

= 28 sq. units

Question 4: Find the area of quadrilateral ABCD whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

Solution:

Given: ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

By Joining AC, we get two triangles ABC and ADC

rs aggarwal class 10 chapter 6 ex c 3

Question 5: If A (-7, 5), B (-6, -7), C (-3, -8) and D (2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Solution:

rs aggarwal class 10 chapter 6 ex c 4

Question 6: Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution:

Given: A triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5)

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC, as shown in the below figure.

rs aggarwal class 10 chapter 6 ex c 5

rs aggarwal class 10 chapter 6 ex c 6

Question 7: A (7, -3), B (5, 3) and C (3, -1) are the vertices of a ∆ABC and AD is its median. Prove that the median AD divides ∆ABC into two triangles of equal areas.

Solution:

rs aggarwal class 10 chapter 6 ex c 7

So,

Area of triangle ABD:

= 1/2(7(3–1) + 5(1+3) + 4(–3–3))

= 1/2(14 + 20–24)

= 1/2(10)

= 5 sq. units …(1)

Area of triangle ACD:

= 1/2(7(–1–1) + 3(1 + 3) + 4(–3 + 1))

= 1/2(–14 + 12–8)

= 1/2(10)

= 5 sq. units ….(2)

From (1) and (2), we conclude that Area of triangle ABD and ACD is equal.

Hence proved.

Question 8: Find the area of ∆ABC with A (1, -4) and midpoints of sides through A being (2, -1) and (0, -1).

Solution:

Given: A ∆ABC with A(1, -4)

rs aggarwal class 10 chapter 6 ex c 8

Question 9: A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.

Solution:

A(6, 1), B(8, 2) and C(9, 4) are the three vertices of a parallelogram ABCD.

E is the midpoint of DC.

Join AE, AC and BD which intersects at O, where O is midpoint of AC.

rs aggarwal class 10 chapter 6 ex c 9

rs aggarwal class 10 chapter 6 ex c 10

Question 10: (i) If the vertices of ∆ABC be A (1, -3), B (4, p) and C (-9, 7) and its area is 15 square units, find the values of p.

(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is

(7/2, y), find the value of y.

Solution:

We know that:

rs aggarwal class 10 chapter 6 ex c 11

Question 11: Find the value of k so that the area of the triangle with vertices A (k + 1, 1), B (4, -3) and C (7, -k) is 6 square units.

Solution:

rs aggarwal class 10 chapter 6 ex c 12

Question 12: For what value of k (k > 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 square units?

Solution:

rs aggarwal class 10 chapter 6 ex c 13

Question 13: Show that the following points are collinear.

(i) A (2, -2), B (-3, 8) and C (-1, 4)

(ii) A (-5, 1), B (5, 5) and C (10, 7)

(iii) A (5, 1), B (1, -1) and C (11, 4)

(iv) A (8, 1), B (3, -4) and C (2, -5)

Solution:

Points are collinear if the area of a triangle is equal to zero.

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

 

(i) A (2, -2), B (-3, 8) and C (-1, 4)

Δ = 1/2{2 (8– 4) + (–3) (4 + 2) –1 (2– 8)}

Δ = 1/2 {8–18 + 10}

Δ = 0

Hence points are collinear.

(ii) A (-5, 1), B (5, 5) and C (10, 7)

Δ = 1/2{–5(5– 7) + 5 (7–1) + 10 (1–5)}

Δ = 1/2{10 + 30–40}

Δ = 0

Hence points are collinear.

(iii) A (5, 1), B (1, -1) and C (11, 4)

Δ = 1/2{5(–1– 4) + 1 (4– 1) + 11 (1 + 1)}

= 1/2{–25 + 3 + 22}

= 0

Hence points are collinear.

(iv) A (8, 1), B (3, -4) and C (2, -5)

Δ = 1/2{8(–4 + 5) + 3 (–5–1) + 2 (1 + 4)}

= 1/2{8–18 + 10}

= 0

Hence points are collinear.

Question 14: Find the value of x for which the points A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Solution:

Points are A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 14

Since points are collinear:

½(x + 63) = 0

Or x = -63

Question 15: For what value of x are the points A (-3, 12), B (7, 6) and C (x, 9) collinear?

Solution:

Points are A (-3, 12), B (7, 6) and C (x, 9) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 15

Since points are collinear:

½(6x -12) = 0

Or x = 2

Question 16: For what value of y are points P (1, 4), Q (3, y) and R (-3, 16) are collinear?

Solution:

Points are P (1, 4), Q (3, y) and R (-3, 16) are collinear.

Which means area of triangle PQR = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 16

Since points are collinear:

½(4y + 8) = 0

Or y = -2

Question 17: Find the value of y for which the points A (-3, 9), B (2, y) and C (4, -5) are collinear.

Solution:

Points are A (-3, 9), B (2, y) and C (4, -5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 17

Since points are collinear:

½(-7y – 7) = 0

Or y = -1

Question 18: For what values of k are the points A (8, 1), B (3, -2k) and C (k, -5) collinear.

Solution:

Points are A (8, 1), B (3, -2k) and C (k, -5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 18

Since points are collinear:

½(2k^2 – 15k + 22) = 0

Or 2k^2 – 15k + 22 = 0

2k^2 – 11k – 4k + 22 = 0

k(2k – 11) – 2(2k – 11) = 0

(k-2)(2k -11) = 0

k = 2 or k = 11/2. Answer.

Question 19: Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.

Solution:

Points are A(2, 1), B(x, y) and C(7, 5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 19

Since points are collinear:

½(4x -5y – 3) = 0

4x – 5y – 3 = 0

Relationship between x and y.

Question 20: Find a relation between x and y, if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Solution:

Points are A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 20

Since points are collinear:

½(2x +y + 3) = 0

2x +y + 3 = 0

Relationship between x and y.

Question 21: Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if 1/a + 1/b = 1.

Solution:

Points are A(a, 0), B(0, b) and C(1, 1) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

= ½[(a(b-1) + 0(1-0) + 1(0-b)]

= ½[ab – a + 0 – b]

Since points are collinear:

½(ab -a-b) = 0

ab – a – b = 0

Divide each term by “ab”, we get

1 – 1/b -1/a = 0

or 1/a + 1/b = 1. hence proved.

Question 22: If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, find the values of a and b.

Solution:

Points are P(-3, 9), Q(a, b) and R(4, -5) are collinear.

Which means area of triangle = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

rs aggarwal class 10 chapter 6 ex c 21

Since points are collinear, we have

½(-14a – 7b + 21) = 0

-14a – 7b + 21 = 0

2a + b =3 …..(1)

a + b = 1 …..(2) (given)

From (1) and (2)

a = 2 and b = -1

Question 23: Find the area of ∆ABC with vertices A (0, -1), B (2, 1) and C (0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.

Solution:

Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

rs aggarwal class 10 chapter 6 ex c 22

Area of triangle DEF:

= 1/2[1+0+1]

= 1 sq. units

Therefore,

Ratio in the area of triangles ABC and DEF = 4/1 = 4:1.

Question 24: If a ≠ b ≠ c, prove that (a, a²), (b, b²), (0, 0) will not be collinear.

Solution:

Let A (a, a²), B (b, b²) and C (0, 0) are the vertices of a triangle.

Let us assume that that points are collinear, then area of ∆ABC must be zero.

Now, area of ∆ABC

rs aggarwal class 10 chapter 6 ex c 23

Which is contraction to our assumption.

This implies points are not be collinear. Hence proved.

R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6C Topics

Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6C is based on following topics:

  • Area of a triangle

Area of ∆ABC whose vertices are is (x_1,y_1 ),(x_2,y_2 )and (x_3,y_3 ) are

Area of a triangle

Students can also find the collinearity of three points using area of triangle formula.